PHYSICS CODE –P2

Section - A

1. If E and G respectively denote energy and gravitational constant, then E

G has the dimensions of:

(1) [M2] [L–2] [T–1] (2) [M2] [L–1] [T0]

(3) [M] [L–1] [T–1] (4) [M] [L0] [T0]

Ans. (2)

E = energy = [ML2T–2]

G = Gravitational constant = [M–1L3T–2]

So, 2 2

2 1 0

1 3 2

E [E] ML T[M L T ]

G [G] M L T

2. An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance ‘R’ are

connected in series to an ac source of potential difference ‘V volts as shown in figure. Potential

difference across L, C and R is 40V, 10V and 40V, respectively. The amplitude of current flowing

through LCR series circuit is 10 2 A. The impedance of the circuit is :

10V 40V 40V

V

(1) 5 (2) 4 2 (3) 5 / 2 (4) 4

Ans. (1)

I0 = 10 2 A

RMS

II 10A

2

10V 40V 40V

V

2 2

RMS R L CV V (V V )

2 2(40) (40 10)

= 50 V

RMS

RM

V 50VZ 5

I 10V

3. A body is executing simple harmonic motion with frequency ‘n’, the frequency of its potential

energy is :

(1) 4n (2) n (3) 2n (4) 3n

Ans. (3)

Displacement equation of SHM of frequency ‘n’

x = A sin (t) = A sin (2nt)

Now,

Potential energy

2 2 21 1U kx KA sin (2 nt)

2 2

21 1 cos(2 (2n)t)kA

2 2

So frequency of potential energy = 2n

4. A thick current carrying cable of radius ‘R’ carries current ‘I’ uniformly distributed across its

cross section. The variation of magnetic field B(r) due to the cable with the distance ‘r’ from the

axis of the cable is represented by:

(1)

r

B (2)

r

B

(3)

r

B (4)

r

B

Ans. (4)

Bin =

0

2

Ir

2 R

B0 =

0I

2 R

r

B

5. A nucleus with mass number 240 breaks into two fragments each of mass number 120, the

binding per nucleon of unfragmented nuclei is 7.6MeV while that of fragments is 8.5MeV. The

total gain in the Binding Energy in the process is :

(1) 216MeV (2) 0.9MeV (3) 9.4MeV (4) 804MeV

Ans. (1)

X240 Y120

Z120

given binding energy per nucleon of X, Y & Z are

7.6 MeV, 8.5 MeV & 8.5 MeV respectively.

Gain in binding energy is:

Q = Binding Energy of products – Binding energy of reactants

= (120 × 8.5 × 2) – (240 × 7.6) MeV

= 216 MeV

6. A parallel plate capacitor has a uniform electric field E in the space between the plates. If the

distance between the plates is ‘d’ and the area of each plate is ‘A’, the energy stored in the

capacitor is: (0 = permittivity of free space)

(1)

2

0

E Ad (2) 2

0

1E

2 (3) 0EAd (4) 2

0

1E Ad

2

Ans. (4)

(4)

Energy = Energy density × volume

= 1

20E

2Ad

7. The number of photons per second on an average emitted by the source of monochromatic light

of wavelength 600nm, when it delivers the power of 3.3 × 10–3 watt will be: (h = 6.6 × 10–34Js)

(1) 1015 (2) 1018 (3) 1017 (4) 1016

Ans. (4)

nhc p

p nhc

3 9

16

34 8

3.3 10 600 10n 10

6.6 10 3 10

8. Polar molecules are the molecules:

(1) having a permanent electric dipole moment

(2) having zero dipole moment

(3) acquire a dipole moment only in the presence of electric field due to displacement of charges

(4) acquire a dipole moment only when magnetic field is absent

Ans. (1)

Polar molecules have centres of postive and negative charges separated by some distance, so they have

permanent dipole moment.

9. The half life of a radioactive nuclide is 100 hours. The fraction of original activity that will

remain after 150 hours would be :

(1) 2

3 2 (2)

1

2 (3)

1

2 2 (4)

2

3

Ans. (3)

Ht/T 150/100

0

A 1 1 1

A 2 2 2 2

10. A capacitor of capacitance ‘C’, is connected across an ac source of voltage V, given by

V = V0sint

The displacement current between the plates of the capacitor, would then be given by :

(1) Id =V0Csint

(2) Id =V0Ccost

(3) Id =

0V

cos tC

(4) Id =

0V

Csint

Ans. (2)

q = CV

dq CdV

dt dt

Id = C(V0 cost)

= V0C cost

11. A screw gauge gives the following readings when used to measure the diameter of a wire

Main scale reading : 0 mm

Circular scale reading : 52 divisions

Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter

of the wire from the above data is :

(1) 0.052 cm (2) 0.52 cm (3) 0.026 cm (4) 0.26 cm

Ans. (1)

L.C. = Pitch

C.SD

=1mm

100 = 0.01m = 0.001cm

Radius = M.S. + n(L–I)

= 0 + 52 (0.001)

= 0.052 cm

12. Find the value of the angle of emergence from the prism. Refractive index of the glass is 3

60°

(1) 90° (2) 60° (3) 30° (4) 45°

Ans. (2)

60°

r2

r1= 3

30°

r1 + r2 = A = 30°

r2 = 30° (r1 = 0°)

from Snell's law

23 sinr 1 sine

13. In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If

another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance

point occurs?

(1) 62 cm (2) 60 cm (3) 21.6 cm (4) 64 cm

Ans. (2)

= constant

Eunknown = lb Eunknown lb

1 1

2 2

E l

E l

1.5 36

2.5 x

x =36 5

3 = 60cm

14. A particle is released from height S from the surface of the Earth. At a certain height its kinetic

energy is three times its potential energy. The height from the surface of earth and the speed of

the particle at that instant are respectively -

(1) S 3gS

,4 2

(2) S 3gS

,4 2

(3) 3gSS

,4 2

(4) 3gSS

,2 2

Ans. (1)

PE + KE = mgs

at given point

KE = 3PE

So, 4PE = mgs

4mgh = mgs

H = s/4

KE = 3mgs 1

4 2mV2

V = 3gs3gs

4 2

15. The effective resistance of a parallel connection that consists of four wires of equal length, equal

area of cross-section and same material is 0.25. What will be the effective resistance if they

are connected in series ?

(1) 4 (2) 0.25 (3) 0.5 (4) 1

Ans. (1)

R

4 = .25 parallel

R = 1

RS = 4R = 4

16. Water falls from a height of 60m at the rate of 15 kg/s to operate a turbine. The losses due to

frictional force are 10% of the input energy. How much power is generated by the turbine?

(g = 10 m/s2)

(1) 7.0 kW (2) 10.2kW (3) 8.1 kW (4) 12.3 kW

Ans. (3)

E = mgh

Pinput = mgh

t

= 15 10 60

1 = 9000 = 9kW

10% loss = 0.9 × 103

Poutput = 9 × 103 – 0.9 × 103 = 8.1 kw

17. An infinitely long straight conductor carries a current of 5A as shown. An electron is moving

with a speed of 105 m/s parallel to the conductor. The perpendicular distance between the

electron and the conductor is 20cm at an instant. Calculate the magnitude of the force

experienced by the electron at that instant.

Electron v = 105 m/s

20cm

P Q 5A

(1) 8 × 10–20N (2) 4 × 10–20N

(3) 8 × 10–20N (4) 4 × 10–20N

Ans. (1)

0I

B2 R

F BVqsin

= 90°

F = BVq

7

5 190

2

I 2 10 5F V e 10 1.6 10

2 R 20 10

20F 8 10 N

18. Consider the following statements (A) and (B) and identify the correct answer.

(A) A Zener diode is connected in reverse bias, when used as a voltage regulator.

(B) The potential barrier of p-n junction lies between 0.1 V to 0.3 V.

(1) (A) is incorrect but (B) is correct.

(2) (A) and (B) both are correct.

(3) (A) and (B) both are incorrect.

(4) (A) is correct and (B) is incorrect.

Ans. (4)

Reverse bias Zener diode use as a voltage regulator

for Ge Potential barrier V0 = 0.3 V

Si Potential barrier V0 = 0.7 V

19. A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass

of 2 kg is suspended by it is -

(1) 0.628 s (2) 0.0628 s (3) 6.28 s (4) 3.14 s

Ans. (1)

F = Kx

10 = K × 0.05

K = 1000

5 = 200

T = m 2

2 2k 200

=

2 6.28

10 10

= .628 s

20. The electron concentration in an n-type semiconductor is the same as hole concentration in a p-

type semiconductor. An external field (electric) is applied across each of them. Compare the

currents in them.

(1) No current will flow in p-type, current will only flow in n-type

(2) Current in n-type = current in p-type

(3) current in p-type > current in n-type

(4) current in n-type > current in p-type.

Ans. (4)

In N type semiconductor majority charge carriersare e– and P type semiconductor majority

chargecarriers are holes.

I = neAVd = neA (E)

e>h = Ie> Ih

21. A dipole is placed in an electric field as shown. In which direction will it move?

+q -q E

(1) towards the right as its potential energy will increase.

(2) towards the left as its potential energy will increase.

(3) towards the right as its potential energy will decrease.

(4) towards the left as its potential energy will decrease.

Ans. (3)

1 2

E E

As field lines are closer at charge +q.

So, net force on the dipole acts towards right side. A system always moves to decrease it’s

potential energy.

1E

2E

E

qE qE

22. A convex lens ‘A’ of focal length 20 cm and a concave lens ‘B’ of focal length 5 cm are kept

along the same axis with a distance ‘d’ between them. If a parallel beam of light falling on ‘A’

leaves ‘B’ as a parallel beam, then the distance ‘d’ in cm will be:

(1) 30 (2) 25 (3) 15 (4) 50

Ans. (3)

d = f1 – f2

= 20 – 5

= 15 cm

23. The escape velocity from the Earth’s surface is . The escape velocity from the surface of

another planet having a radius, four times that of Earth and same mass density is:

(1) 4 (2) (3) 2 (4) 3

Ans. (1)

ve = 2GM

R = 32G 4

RR 3

= 38 GR

3

ve R

ev

=

4R

R ve = 4v

24. An electromagnetic wave of wavelength ‘’ is incident on a photosensitive surface of negligible

work function. If ‘m’ mass is of photoelectron emitted from the surface has de-Broglie

wavelength d, then:

(1) 2

d

2h

mc

(2) 2

d

2m

hc

(3) 2

d

2mc

h

(4) 2

d

2mc

h

Ans. (4)

hc

= kmax + [given is negligible]

So, hc

= Kmax

d =

max

h

2mK Kmax =

2

2

d

h

2m

hc

= 2

2

d

h

2m = 2

d

2mc

h

25. A lens of large focal length and large aperture is best suited as an objective of an astronomical

telescope since:

(1) a large aperture contributes to the quality and visibility of the images.

(2) a large area of the objective ensures better light gathering power.

(3) a large aperture provides a better resolution.

(4) All the above.

Ans. (4)

MP = 0

e

f

f

R.P. = a

1.22

Large aperture(a) of the objective lens provides bettern resolution

good quality of image is formed and also it gathers more light.

26. Two charged spherical conductors of radius R1 and R2 are connected by a wire. Then the ratio of

surface charge densities of the spheres (1/ 2) is:

(1) 2

1

2

2

R

R (2) 1

2

R

R (3) 2

1

R

R (4)

1

2

R

R

Ans. (3)

R1 R2

Q1 =

1

1 2

QR

R R

Q2 =

2

1 2

QR

R R

1 =

1 1

2 2

1 2 11 1

QQ R 1

R R R4 R 4 R

2 =

2 2

2 2

1 2 22 2

QQ R 1

R R R4 R 4 R

1 2

2 1

R

R

27. For a plane electromagnetic wave propagating in x-direction, which one of the following

combination gives the correct possible directions for electric field (E) and magnetic field (B)

respectively?

(1) ˆ ˆˆ ˆj k, j k (2) ˆ ˆˆ ˆj k, j k

(3) ˆ ˆˆ ˆj k, j k (4) ˆ ˆˆ ˆj k, j k

Ans. (3)

Wave in x direction

C = E B

ˆ ˆˆ ˆ– j k – j k

= i + i

= 2i

28. A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The

time taken by a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at

20°C is:

(1) 5

t13

(2) 13

t10

(3) 13

t5

(4) 10

t13

Ans. (3)

dT

dt=K(Tav – T0)

10

t= K(85 – 20)

20

t = K(70 – 20)

t ' 65

2t 50

t ' 130

t 50

13

t ' t5

29. The equivalent capacitance of the combination shown in the figure is:

C

C

C

(1) 3C/2 (2) 3C (3) 2C (4) C/2

Ans. (3)

C B

B C

A B

C

C

A B

CAB = 2C

30. If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities.

Find the dimensions of energy.

(1) [F] [A–1] [T]

(2) [F] [A] [T]

(3) [F] [A] [T2]

(4) [F] [A] [T–1]

Ans. (3)

E Fa Ab Tc

[M1L2T–2] [M1L1T–2]a [LT–2]b [T]c

a = 1

a + b = 2 b = 1

–2a – 2b + c = –2

c = 2

a = 1 b = 1 c = 2

E [F] [A] [T2]

31. A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let Sn be

the distance travelled by the block in the interval t = n – 1 to t = n. Then the ratio

n

n 1

S

S is :

(1) 2n

2n–1 (2)

2n–1

2n (3)

2n–1

2n 1 (4)

2n 1

2n–1

Ans. (3)

n

n 1

a2n–1S 2n–1 2n–12

aS 2n 2 –1 2n 12 n 1 –1

2

32. The velocity of a small ball of mass M and density d, when dropped in a container filled with

glycerine becomes constant after some time. If the density of glycerine is d

2, then the viscous

force acting on the ball will be :

(1) 2Mg (2) Mg

2 (3) Mg (4)

3

2Mg

Ans. (2)

w Mg vdy

B

d MgF M'g v y

2 2

Fv + FB = Fg

V

mg mgF mg

2 2

33. Column-I gives certain physical terms associated with flow of current through a metallic

conductor. Column-II gives some mathematical relations involving electrical quantities. Match

column-I and column-II with appropriate relations.

Column-I Column-II

(A) Drift Velocity (P) 2

m

ne

(B) Electrical Resistivity (Q) ned

(C) Relaxation Period (R) eE

m

(D) Current Density (S) E

J

(1) (A)-(R); (B)-(Q); (C)-(S); (D)-(P)

(2) (A)-(R); (B)-(S); (C)-(P); (D)-(Q)

(3) (A)-(R); (B)-(S); (C)-(Q); (D)-(P)

(4) (A)-(R); (B)-(P); (C)-(S); (D)-(Q)

Ans. (2)

J = I

A = nevd =

2ne

mE = E =

E

Vd = eE

Im

; = 2

m

ne or =

E

J, J = nevd

A R B S D Q

= 2

m

ne

C P

34. A radioactive nucleus A

ZX undergoes spontaneous decay in the sequence

A

Z Z 1 Z 3 Z 2X B C D

, Where Z is the atomic number of element X.

The possible decay particles in the sequence are:

(1) –, ,

(2) –, ,

(3) –, ,

(4) –, ,

Ans. (4)

+ decreases atomic number by 1

decreases atomic number by 2

– increases atomic number by 1

35. Match Column-I and Column-II and choose the correct match from the given choices.

Column-I Column-II

(A) Root mean square speed of gas molecules (P) 21nmv

3

(B) Pressure exerted by ideal gas (Q) 3RT

M

(C) Average kinetic energy of a molecule (R) 5

RT2

(D) Total internal energy of 1mole of a diatomic gas (S) B

3k T

2

(1) (A) – (R), (B) – (Q), (C) – (P), (D) – (S)

(2) (A) – (R), (B) – (P), (C) – (S), (D) – (Q)

(3) (A) – (Q), (B) – (R), (C) – (S), (D) – (P)

(4) (A) – (Q), (B) – (P), (C) – (S), (D) – (R)

Ans. (4)

(A)rms

3RTV

M

(B) 2

m

1P nmV

3

(C) 3

E kT2

(D) Total

f 5E n RT RT

2 2

Section - B

36. In the product

F q v B

=

0ˆˆ ˆqv Bi Bj B k

For q = 1 and ˆˆ ˆv 2i 4j 6k and

ˆˆ ˆF 4i –20j 12k

What will be the complete expression for B ?

(1) ˆˆ ˆ6i 6j – 8k (2) ˆˆ ˆ–8i – 8j – 6k

(3) ˆˆ ˆ–6i – 6j – 8k (4) ˆˆ ˆ8i 8j – 6k

Ans. (3)

F q(V B)

w

By check options

(C) ˆˆ ˆ6i 6j 8k

i j k

2 4 6

6 6 8

ˆ ˆV B i( 32 36) j( 36 16)

+ K(–12 + 24)

ˆˆ ˆ4i 20j 12k

37. From a circular ring of mass ‘M’ and radius ‘R’ an arc corresponding to a 90° sector is removed.

The moment of inertia of the remaining part of the ring about an axis passing through the

center of the ring and perpendicular to the plane of the ring is ‘K’ times ‘MR2’. Then the value of

‘K’ is:

(1) 1

8 (2)

3

4 (3)

7

8 (4)

1

4

Ans. (2)

I

I I

M/4

M/4 M/4

I = MR2

I’ = 23MR

4

I’ = 3

I4

38. Three resistors having resistances r1, r2 and r3 are connected as shown in the given circuit. The

ratio 3

1

i

i of currents in terms of resistances used in the circuit is:

i2

B

r2

i3 r3

r1

i1 A

(1)

2

1 3

r

r r (2)

1

2 3

r

r r (3)

2

2 3

r

r r (4)

1

1 2

r

r r

Ans. (3)

I3 = 1 2

2 3

I r

r r

3 1 2 2

1 2 32 3 1

I I r r

I r rr r I

39. A ball of mass 0.15 kg is dropped from a height 10m, strikes the ground and rebounds to the

same height. The magnitude of impulse imparted to the ball is (g = 10 m/s2) nearly:

(1) 1.4 kg m/s (2) 0 kg m/s

(3) 4.2 kg m/s (4) 2.1 kg m/s

Ans. (3)

V = 2gh

V = 2 10 10

V = 10 2

I = 2mV

= 2 × 1.5 × 10 2

= 3 2

= 3 × 1.4 = 4.2 kgm/s

40. A step down transformer connected to an ac mains supply of 220 V is made to operate at 11 V,

44W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit.

(1) 4A (2) 0.2A (3) 0.4 A (4) 2A

Ans. (2)

Power loss = 0

= 100% Pin = Po/p

VPIP = VSIs

220 × IP = 44

IP = 44 1

220 5A = .2A

41. A uniform conducting wire of length 12a and resistance ‘R’ is wound up as a current carrying

coil in the shape of,

(i) an equilateral triangle of side ‘a’

(ii) A square of side ‘a’

The magnetic dipole moments of the coil in each case respectively are:

(1) 4Ia2 and 3Ia2 (2) 2 23 Ia and 3I a

(3) 3Ia2 and Ia2 (4) 3Ia2 and 4Ia

2

Ans. (2)

x = 2 2 2

2 a 3a 3a 3a – a

4 4 4 2

A1 = 1 3

a a2 2

A1 = 23a

4

1 = N1IA1

1= 24I 3a

4

1 = 23Ia

a a

a

a

N2=3

A2 = a2

2 = N2IA2

= 3 × I × a2

2 = 3Ia2

42. A particle of mass ‘m’ is projected with a velocity = kVe (k < 1) from the surface of the earth.

(Ve = escape velocity)

The maximum height above the surface reached by the particle is:

(1)

2

2

Rk

1 k (2)

2

2

kR

1 k (3)

2

2

kR

1 k (4)

2R k

1 k

Ans. (1)

2 2

e

GMm 1 GMmmk v

R 2 r

2GMm 1 2GM GMmmk

R 2 R r

21 k 1

R R r

21 1 k

r R R

21 1 k

r R

2

Rr

1 k

2

RR h

1 k

2

Rh R

1 k

2

2

kR

1 k

43. A series LCR circuit containing 5.0 H inductor, 80F capacitor and 40 resistor is connected to

230V variable frequency ac source. The angular frequencies of the source at which power

transferred to the circuit is half the power at the resonant angular frequency are likely to be:

(1) 42 rad/s and 58 rad/s (2) 25 rad/s and 75 rad/s

(3) 50 rad/s and 25 rad/s (4) 46 rad/s and 54 rad/s

Ans. (4)

L 50

Q R / L 8 rad / secR 4

0

6

1 150 rad / sec.

LC 5 80 10

min 0

462

rad/sec

max 0

542

rad /sec

44. A particle moving in a circle of radius R with a uniform speed takes a time T to complete one

revolution. If this particle were projected with the same speed at an angle ‘’ to the horizontal,

the maximum height attained by it equals 4R. The angle of projection, ,is then given by :

(1) = sin–1

1/22

2

2gT

R (2) = cos–1

1/22

2

gT

R

(3) = cos–1

1/22

2

R

gT (4) = sin–1

1/22

2

R

gT

Ans. (1)

T = 2 R

V

V = 2 R

T

H = 2 2u sin

2g

4R = 2 2 2

2

4 R sin

T 2g

sin2 =

2

2 2

8RT g

4 R

sin =

2

2

2T g

R

= sin–1

1/22

2

2T g

R

45. Two conducting circular loops of radii R1 and R2 are placed in the same plane with their centres coinciding. If R1>> R2, the mutual inductance M between them will be directly proportional to:

(1) 2

2

1

R

R (2) 1

2

R

R (3) 2

1

R

R (4)

2

1

2

R

R

Ans. (1)

M =

2 2

1

N

I

M = 2 1 2

1

N B A

I

M B1 A2

M

0 1

1

I

2 R×

2

2R

M 2

2

1

R

R

46. Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger

drop. Calculate the potential of the bigger drop. (1) 1980 V (2) 660 V (3) 1320 V (4) 1520 V Ans. (1) VB = N2/3VS VB = (27)2/3 .(220) VB = 9 × 220 = 1980V 47. A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark.

A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass ‘m’ is suspended from the rod at 160 cm mark as shown in the figure. Find the value of ‘m’ such that the rod is in equilibrium. (g = 10 m/s2)

0 20cm 40cm 160cm

2kg m

(1) 1

12kg (2)

1

2kg (3)

1

3kg (4)

1

6kg

Ans. (1) By balancing torque 2g × 20 = 0.5g × 60 + mg × 120

m = 0.5

6 kg =

1

12 kg

48. A point object is placed at a distance of 60 cm from a convex lens of focal length 30 cm. If a

plane mirror were put perpendicular to the principal axis of the lens and at a distance of 40 cm

from it, the final image would be formed at a distance of –

60cm 40cm (1) 20 cm from the plane mirror, it would be a virtual image.

(2) 20 cm from the lens, it would be a real image

(3) 30 cm from the lens, it would be a real image

(4) 30 cm from the plane mirror, it would be a virtual image.

Ans. (1)

First for image formation from lens

u = –60cm

f = +30cm

v = uf

u f =

–60 30

–60 30

= 60cm

This real image formed by lens acts as virtual object for mirror

Real image from plane mirror is formed 20cm in front of mirror, hence at 20cm distance from

lens. Now for second refraction from lens,

u = –20cm

f = +30cm

v = uf –20 30

u f –20 30

= –60cm

So, final virtual image is 60cm from lens, or 20cm behind mirror

I3

20cm 20cm

O1

60cm

49. A car starts from rest and accelerates at 5m/s2. At t=4 s, a ball is dropped out of a window by a

person sitting in the car. What is the velocity and acceleration of the ball at t=6 s ?

(Take g = 10 m/s2)

(1) 20 2 m/s, 10m/s2

(2) 20m/s, 5m/s2

(3) 20m/s, 0

(4) 20 2 m/s, 0

Ans. (1)

u = 0

a = 5

t = 4 V = u + at

V = 0 + 5 × 4

V = 20

Vx = 20 m/sec

Vy = u + ut

= 10 × 2

Vy= 20m/sec

V = 20 2

and a = 10 m/sec2

50. For the given circuit, the input digital signals are applied at the terminals A, B and C. What

would be the output at the terminal y?

A

5

0

5

0

0

5

t1 t2 t3 t4 t5 t6

B

C

y

A

B

C

(1)

t1 t2 t3 t4 t5 t6

y

0V

(2)

0V

5V

(3)

5V

(4)

0V

5V

Ans. (3)

AB

BC

Y = AB + BC

A B C Y

0 0 1 0 1 1

1 0 1 0 1 1

0 1 0 0 1 1

0 0 1 0 1 1